LeetCode Populating Next Right Pointers in Each Node II(dfs)
问题：给出一个二叉查找树，将结点与其右边的结点相连
思路：
从顶向上，从右向左的方式 。递归过程中，在当前结点及父结点作为参数传递。在向下的过程中，如果父结点不为空则获取当前结点的next结点。如果当前结点是左孩子，并且右孩子不为空，则说明当前结点的next是其父结点的右孩子。如果不是，则查看父结点的next结点，如果其有左孩子，说明当前结点的next就是上一层的其中的一个左孩子，否则就中右孩子。
具体代码参考：
https://github.com/wuli2496/OJ/tree/master/LeetCode/Populating%20Next%20Right%20Pointers%20in%20Each%20Node%20II
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